Difference between revisions of "2022 AIME II Problems/Problem 4"
m (→Solution 1) |
|||
Line 3: | Line 3: | ||
There is a positive real number <math>x</math> not equal to either <math>\tfrac{1}{20}</math> or <math>\tfrac{1}{2}</math> such that<cmath>\log_{20x} (22x)=\log_{2x} (202x).</cmath>The value <math>\log_{20x} (22x)</math> can be written as <math>\log_{10} (\tfrac{m}{n})</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | There is a positive real number <math>x</math> not equal to either <math>\tfrac{1}{20}</math> or <math>\tfrac{1}{2}</math> such that<cmath>\log_{20x} (22x)=\log_{2x} (202x).</cmath>The value <math>\log_{20x} (22x)</math> can be written as <math>\log_{10} (\tfrac{m}{n})</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
==Solution 1== | ==Solution 1== | ||
− | Define <math>a</math> to be <math>\log_{20x} (22x) = \log_{2x} (202x)</math>, what we are looking for. Then, by the definition of | + | Define <math>a</math> to be <math>\log_{20x} (22x) = \log_{2x} (202x)</math>, what we are looking for. Then, by the definition of the logarithm, |
<cmath>\begin{cases} | <cmath>\begin{cases} | ||
(20x)^{a} &= 22x \ | (20x)^{a} &= 22x \ | ||
Line 11: | Line 11: | ||
~ihatemath123 | ~ihatemath123 | ||
+ | |||
==Solution 2== | ==Solution 2== | ||
We could assume a variable <math>v</math> which equals to both <math>\log_{20x} (22x)</math> and <math>\log_{2x} (202x)</math>. | We could assume a variable <math>v</math> which equals to both <math>\log_{20x} (22x)</math> and <math>\log_{2x} (202x)</math>. |
Revision as of 01:10, 13 June 2022
Contents
[hide]Problem
There is a positive real number not equal to either or such thatThe value can be written as , where and are relatively prime positive integers. Find .
Solution 1
Define to be , what we are looking for. Then, by the definition of the logarithm, Dividing the first equation by the second equation gives us , so by the definition of logs, . This is what the problem asked for, so the fraction gives us .
~ihatemath123
Solution 2
We could assume a variable which equals to both and .
So that and
Express as:
Substitute to :
Thus, , where and .
Therefore, .
~DSAERF-CALMIT (https://binaryphi.site)
Solution 3
We have
We have
Because , we get
We denote this common value as .
By solving the equality , we get .
By solving the equality , we get .
By equating these two equations, we get
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4
By the change of base rule, we have , or . We also know that if , then this also equals . We use this identity and find that . The requested sum is
~MathIsFun286
Video Solution
https://www.youtube.com/watch?v=4qJyvyZN630
Video Solution by Power of Logic
~Hayabusa1
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.